On the Kantorovich Theory for Nonsingular and Singular Equations

Abstract

We develop a new Kantorovich-like convergence analysis of Newton-type methods to solve nonsingular and singular nonlinear equations in Banach spaces. The outer or generalized inverses are exchanged by a finite sum of linear operators making the implementation of these methods easier than in earlier studies. The analysis uses relaxed generalized continuity of the derivatives of operators involved required to control the derivative and on real majorizing sequences. The same approach can also be implemented on other iterative methods with inverses. The examples complement the theory by verifying the convergence conditions and demonstrating the performance of the methods.

1. Introduction

Let $T_1,T_2$ denote Banach spaces, and let $\mathcal{B}(T_1,T_2)$ be the space of linear and continuous operators from $T_1$ to $T_2.$ Newton-type methods (NTMs) [1]

$$\begin{array}{c}\hfill x_{n+1}=x_n-E{\left(x_n\right)}^{\#}\mathrm{{\rm Y}}\left(x_n\right),n=0,1,2,\dots \end{array}$$

(1)

have been used to solve the equation

$$\begin{array}{c}\hfill D\mathrm{{\rm Y}}\left(x\right)=0.\end{array}$$

(2)

Here, the operator $\mathrm{{\rm Y}}:\mathrm{\Omega }\subset T_1⟶T_2$ is a differentiable operator in the Fréchet sense, $D\in \mathcal{B}(T_2,T_1),E\left(x_n\right)\in \mathcal{B}(T_1,T_2)$ approximates ${\mathrm{{\rm Y}}}^{\prime }\left(x_n\right).$ Moreover, $E{\left(x_n\right)}^{\#}$ stands for an outer inverse (OI) of $E\left(x_n\right)$, i.e., $E{\left(x_n\right)}^{\#}E\left(x_n\right)E{\left(x_n\right)}^{\#}=E{\left(x_n\right)}^{\#}.$

A plethora of applications in optimization such as penalization problems, minimax problems, and goal programming are formulated as (2) using Mathematical Modelling [2,3,4,5,6,7,8,9,10,11,12,13,14,15].

Method (2) specializes to the Gauss–Newton method (GNM) for solving nonlinear least squares problems, the generalized NTM for undetermined systems, and an NTM for ill-posed Equations [16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36].

As an example of (1) and (2), let $T_1$ and $T_2$ stand for Hilbert spaces. Then, consider the task of finding a local minimum $\tilde{u}$ of

$$minH\left(u\right),$$

(3)

where $H\left(u\right)=\frac{1}{2}{\parallel \mathrm{{\rm Y}}\left(u\right)\parallel }^2.$ Then, the GNM is defined by

$$x_{n+1}=x_n-{\mathrm{{\rm Y}}}^{\prime }{\left(x_n\right)}^{†}\mathrm{{\rm Y}}\left(x_n\right)$$

(4)

to solve

$${\mathrm{{\rm Y}}}^{\prime }{\left(u\right)}^{\ast }\mathrm{{\rm Y}}\left(u\right)=0,$$

(5)

where ${\mathrm{{\rm Y}}}^{\prime }{\left(x_n\right)}^{†}$ is the Moore–Penrose inverse [5,15,31], and ${\mathrm{{\rm Y}}}^{\prime }{\left(u\right)}^{\ast }$ is the adjoint of the linear operator ${\mathrm{{\rm Y}}}^{\prime }\left(u\right)$ (see also the Remark 1).

Ben-Israel [5,17] utilized the conditions

$$\begin{array}{ccc}\hfill \parallel \mathrm{{\rm Y}}\left(u\right)-\mathrm{{\rm Y}}\left(\tilde{u}\right)-{\mathrm{{\rm Y}}}^{\prime }\left(\tilde{u}\right)(u-\tilde{u})\parallel & \le & c_1\parallel u-\tilde{u}\parallel ,\hfill \\ \hfill \parallel \left({\mathrm{{\rm Y}}}^{\prime }{\left(u\right)}^{+}-{\mathrm{{\rm Y}}}^{\prime }{\left(\tilde{u}\right)}^{+}\right)\mathrm{{\rm Y}}\left(\tilde{u}\right)\parallel & \le & c_2\parallel u-\tilde{u}\parallel \hfill \end{array}$$

and

$$\begin{array}{c}\hfill c_1\parallel {\mathrm{{\rm Y}}}^{\prime }{\left(\tilde{u}\right)}^{+}\parallel +c_2<1\end{array}$$

for all $u,\tilde{u}$ in a neighborhood of $x_0\in \mathrm{\Omega }$. He also used these conditions with ${\mathrm{{\rm Y}}}^{\prime }{\left(u\right)}^{\#}$ [5,17]. These results are not semilocal since they require information about ${\mathrm{{\rm Y}}}^{\prime }{\left(\tilde{u}\right)}^{+}$ or ${\mathrm{{\rm Y}}}^{\prime }{\left(\tilde{u}\right)}^{\#}$. Moreover, if ${\mathrm{{\rm Y}}}^{\prime }{\left(x_0\right)}^{-1}\in \mathcal{B}(T_2,T_1)$, they require conditions not required in the Kantorovich theory [1,16,33,37,38,39,40,41]. Later Deuflhard and Heindl [39], Haussler [29], and Yamamoto [1] gave Kantorovich-type theorems for the GNM like (4) using convergence conditions involving either OI of Moore–Penrose inverses:

$$\parallel {\mathrm{{\rm Y}}}^{\prime }{\left(\tilde{u}\right)}^{\#}\left(I-{\mathrm{{\rm Y}}}^{\prime }\left(u\right){\mathrm{{\rm Y}}}^{\prime }{\left(u\right)}^{\#}\right)\mathrm{{\rm Y}}\left(u\right)\parallel \le c\left(u\right)\parallel u-\tilde{u}\parallel ,c\left(u\right)\le \overline{c}<1,$$

(6)

for each $u,\tilde{u}\in \mathrm{\Omega }$. This condition is strong and does not hold in concrete examples (see Section 4 in [31]). A Kantorovich-like result with generalized inverses can be found in [42] without (6). However, it was assumed that $T_1$ and $T_2$ are finite-dimensional and $T_2=R(E\left(x_0\right)$, where $R\left(D\right)$ denotes the range of a linear operator D. Other drawbacks of the earlier works are that only properties of OI are used. That is $BEB=B$ and the projectional properties of $EB$ and $BE$. However, the stability and perturbation bounds for OI are not given for method (1). However, this was accomplished through the elegant work of Nashed and Chen [31]. This work reduces to the Kantorovich theory for (2) when $E{\left(u\right)}^{\#}$ is replaced by $E{\left(u\right)}^{-1}$ without additional conditions. Later works on the convergence analysis using Lipchit-type conditions, particularly for the Newton–Gauss method (5), can be found in [10,24] and the references therein.

Next, we address the problems with the implementation of method (1) which constitutes the motivation for this paper. Let $\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$ and let ${\mathsf{\Delta }}^{\#}\in \mathcal{B}(T_2,T_1)$ be an OI of $\mathsf{\Delta }$.

Suppose ${\mathsf{\Delta }}_1^{\#}\left(u\right)={\left(I+{\mathsf{\Delta }}^{\#}\left(\mathsf{\Delta }-E\left(u\right)\right)\right)}^{-1}{\mathsf{\Delta }}^{\#}$ is an OI. A criterion for ${\mathsf{\Delta }}_1^{\#}$ to be an OI is $\parallel {\mathsf{\Delta }}^{\#}\left(E\left(u\right)-\mathsf{\Delta }\right)\parallel \le 1$ (see Lemma 2). Then, (2) becomes

$$x_{n+1}=x_n-{\mathsf{\Delta }}_1^{\#}\left(x_n\right)\mathrm{{\rm Y}}\left(x_n\right).$$

(7)

But the main problem with the implementation of (2) of (7) still remains. This problem requires the invertibility of ${\mathsf{\Delta }}_2,{\mathsf{\Delta }}_2\left(u\right)=I+{\mathsf{\Delta }}^{\#}\left(\mathsf{\Delta }-E\left(u\right)\right).$ This inversion can be avoided. Let m be a fixed natural number. Define the operators

$$H=H\left(u\right)={\mathsf{\Delta }}^{\#}\left(\mathsf{\Delta }-E\left(u\right)\right)$$

and

$$L=L_m\left(u\right)=I+H+H^2+\cdots +H^m.$$

Then, we can consider the replacement of (7) given as

$$x_{n+1}=x_n-L{\mathsf{\Delta }}^{\#}\mathrm{{\rm Y}}\left(x_n\right).$$

(8)

By letting $m\to +\infty$ in the definition of L, we have that

$$E{\left(x_n\right)}^{\#}=\underset{m\to +\infty }{lim}{L}_m\left(x_m\right){\mathsf{\Delta }}^{\#}.$$

Thus, it is worth studying the convergence of (8) instead of (2), since we avoid the inversion of the operator ${\mathsf{\Delta }}_2$. Let us provide examples of possible choices for the operator $\mathsf{\Delta }$. First, consider the case when the operator $E={\mathrm{{\rm Y}}}^{\prime }\left(u\right)$ is invertible. Moreover, let $T_1=T_2={\mathbb{R}}^i,i$ is a positive integer, and J denotes the Jacobian of the operator F. Then, choose $\mathsf{\Delta }=J\left(x_0\right)$ in the semi-local convergence case of $\mathsf{\Delta }=J\left(x^{\ast }\right)$ in the local case, where $x^{\ast }\in \mathrm{\Omega }$ is assumed to be a solution of the equation $\mathrm{{\rm Y}}\left(u\right)=0.$ The selection $\mathsf{\Delta }=J\left(\overline{u}\right)$ has been used in [43,44], for $\overline{u}\in \mathrm{\Omega }.$ In the setting of a Banach space for $E={\mathrm{{\rm Y}}}^{\prime }\left(u\right),$ the operator $\mathsf{\Delta }$ can be chosen to be $\mathsf{\Delta }={\mathrm{{\rm Y}}}^{\prime }\left(x_n\right)$ (semi-local case) or $\mathsf{\Delta }={\mathrm{{\rm Y}}}^{\prime }\left(x^{\ast }\right)$ (local case). Numerous selections for $\mathsf{\Delta }$ connected to OI or generalized inverses (GI) can also be found in [5,11,13,31] and the references therein. Other selections for $\mathsf{\Delta }$ are also possible provided that they satisfy the convergence conditions ($C_5$) and ($C_6$) of Section 3. The convergence analysis relies on the relaxed generalized continuity used to control the derivative ${\mathrm{{\rm Y}}}^{\prime }$ and majorizing sequences for the iterates $\left\{x_n\right\}$ (see also Section 2). The results in this article specialize immediately to solve nonsingular equations if $E{\left(u\right)}^{\#}$ is replaced by $E{\left(u\right)}^{-1}.$

The rest of the article provides the preliminaries in Section 2; the convergence of (8) is in Section 3; and the applications are in Section 4. The article’s concluding remarks appear in Section 5.

2. Preliminaries

We reproduce standard results on OI and GI to make the article as self-contained as possible. More properties can be found in [5,11,12,13,31]. Let $\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$. An operator $B\in \mathcal{B}(T_2,T_1)$ is said to be an inner inverse (II) of $\mathsf{\Delta }$ if $\mathsf{\Delta }B\mathsf{\Delta }=\mathsf{\Delta }$, and an OI of $\mathsf{\Delta }$ if $B\mathsf{\Delta }B=B$. It is well known that II and bounded OI always exist. The zero is always an OI. So, we consider only nonzero outer inverses. Suppose the operator B is either an inverse or an OI of $\mathsf{\Delta }$. Then, $\mathsf{\Delta }B$ and $B\mathsf{\Delta }$ are linear idempotents (algebraic projectors). Suppose that B is an inverse of E, then for $N\left(\overline{E}\right)$, $N\left(B\mathsf{\Delta }\right)=N\left(\mathsf{\Delta }\right)$ and $R\left(\mathsf{\Delta }\right)=R\left(\mathsf{\Delta }B\right)$. Consequently, the following decompositions hold $T_1=N\left(\mathsf{\Delta }\right)+R\left(B\mathsf{\Delta }\right)$ and $T_2=R\left(\mathsf{\Delta }\right)+N\left(\mathsf{\Delta }B\right)$. Since, B is an OI of $\mathsf{\Delta }$ if and only if $\mathsf{\Delta }$ is an inverse of B, it follows that $T_1=R\left(B\right)\oplus N\left(B\mathsf{\Delta }\right)$ and $T_2=N\left(B\right)\oplus R\left(\mathsf{\Delta }B\right)$. If B is an inner and an outer inverse of $\mathsf{\Delta }$, then B is called a GI of $\mathsf{\Delta }$. Moreover, there exists a unique GI $B={\mathsf{\Delta }}_{P,Q}^{+}$ satisfying $\mathsf{\Delta }B\mathsf{\Delta }=\mathsf{\Delta },$ $B\mathsf{\Delta }B=B,B\mathsf{\Delta }=I-P$ and $\mathsf{\Delta }B=Q,$ where P is a given projection of $T_1$ into $N\left(\mathsf{\Delta }\right)$ and Q a given projection of $T_2$ into $R\left(\mathsf{\Delta }\right)$. In the special case when $T_1$ and $T_2$ are Hilbert spaces, and $P,Q$ are orthogonal projections. Then, ${\mathsf{\Delta }}_{P,Q}^{+}$ is the Moore–Penrose inverse of $\mathsf{\Delta }$.

We need the following six auxiliary Lemmas, the proof of which can be found in Section 2 in [31].

Lemma 1. 

Let $\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$. If ${\mathsf{\Delta }}^{\#}$ is a bounded OI of $\mathsf{\Delta }.$ Then, the following assertions hold

$$T_1=R\left({\mathsf{\Delta }}^{\#}\right)\oplus N\left({\mathsf{\Delta }}^{\#}\mathsf{\Delta }\right)$$

and

$$T_2=N\left({\mathsf{\Delta }}^{\#}\right)\oplus R\left(\mathsf{\Delta }{\mathsf{\Delta }}^{\#}\right).$$

Lemma 2 

(Banach Perturbation-Like Lemma). Let $\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$ and ${\mathsf{\Delta }}^{\#}\in \mathcal{B}(T_2,T_1)$ be an OI of Δ. Let also $B\in \mathcal{B}(T_1,T_2)$ be such that $\parallel {\mathsf{\Delta }}^{\#}(B-\mathsf{\Delta })\parallel <1$. Then, $B^{\#}:={\left(I+{\mathsf{\Delta }}^{\#}(B-\mathsf{\Delta })\right)}^{-1}{\mathsf{\Delta }}^{\#}$ is a bounded OI of B so that $N\left(B^{\#}\right)=N\left({\mathsf{\Delta }}^{\#}\right)$, $R\left(B^{\#}\right)=R\left({\mathsf{\Delta }}^{\#}\right),$

$$\begin{array}{ccc}\hfill \parallel B^{\#}-{\mathsf{\Delta }}^{\#}\parallel & \le & \frac{\parallel {\mathsf{\Delta }}^{\#}(B-\mathsf{\Delta }){\mathsf{\Delta }}^{\#}\parallel }{1-\parallel {\mathsf{\Delta }}^{\#}(B-\mathsf{\Delta })\parallel }\hfill \\ & \le & \frac{\parallel {\mathsf{\Delta }}^{\#}(B-\mathsf{\Delta })\parallel \parallel {\mathsf{\Delta }}^{\#}\parallel }{1-\parallel {\mathsf{\Delta }}^{\#}(B-\mathsf{\Delta })\parallel }\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel B^{\#}\mathsf{\Delta }\parallel & \le & \frac{1}{1-\parallel {\mathsf{\Delta }}^{\#}(B-\mathsf{\Delta })\parallel }.\hfill \end{array}$$

Lemma 3. 

Let $\mathsf{\Delta },B\in \mathbb{B}(T_1,T_2)$ and let ${\mathsf{\Delta }}^{\#}$ and $B^{\#}\in \mathbb{B}(T_2,T_1)$ be OI of Δ, and B, respectively. Then, $B^{\#}(I-\mathsf{\Delta }{\mathsf{\Delta }}^{\#})=0\iff N\left({\mathsf{\Delta }}^{\#}\right)\subset N\left(B^{\#}\right)$.

Lemma 4. 

Let $\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$. Suppose that $T_1$ and $T_2$ have topological decompositions $T_1=N\left(\mathsf{\Delta }\right)\oplus M,$$T_2=R\left(\mathsf{\Delta }\right)\oplus S$. Let ${\mathsf{\Delta }}^{+}(={\mathsf{\Delta }}_{M,S}^{+})$ stand for the GI of Δ connected to these decompositions. Let B satisfy $\parallel {\mathsf{\Delta }}^{+}(B-\mathsf{\Delta })\parallel <1$, and $\left(I+(B-\mathsf{\Delta }){\mathsf{\Delta }}^{+}\right)B$ sends $N\left(\mathsf{\Delta }\right)$ to $R\left(\mathsf{\Delta }\right)$. Then, the following assertions hold:

$$B^{+}:=B_{R\left({\mathsf{\Delta }}^{+}\right),N\left({\mathsf{\Delta }}^{+}\right)}^{+}\mathit{exists},$$

$$B^{+}={\mathsf{\Delta }}^{+}{(I+B_0{\mathsf{\Delta }}^{+})}^{-1}={(I+{\mathsf{\Delta }}^{+}{B}_0)}^{-1}{\mathsf{\Delta }}^{+},$$

$$R\left(B^{+}\right)=R\left({\mathsf{\Delta }}^{+}\right),N\left(B^{+}\right)=N\left({\mathsf{\Delta }}^{+}\right)$$

and

$$\parallel B^{+}\mathsf{\Delta }\parallel \le \frac{1}{\parallel {\mathsf{\Delta }}^{+}(B-\mathsf{\Delta })\parallel },$$

where $B_0=B-\mathsf{\Delta }$.

Lemma 5. 

Let $\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$ and ${\mathsf{\Delta }}^{+}$ be the GI as given in the Lemma 4. Let $B\in \mathcal{B}(T_2,T_1)$ satisfy $\parallel {\mathsf{\Delta }}^{\#}(B-\mathsf{\Delta })\parallel \le 1$ and $R\left(B\right)\subseteq R\left(\mathsf{\Delta }\right)$. Then, the conclusions of the Lemma 4 hold and $R\left(B\right)=R\left(\mathsf{\Delta }\right)$.

Lemma 6. 

Let $\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$ and let ${\mathsf{\Delta }}^{+}$ be bounded GI of Δ. Let $B\in \mathcal{B}(T_2,T_1)$ satisfy $\parallel {\mathsf{\Delta }}^{+}(B-\mathsf{\Delta })\parallel <1$. Let $B^{\#}:={\left(I+{\mathsf{\Delta }}^{+}(B-\mathsf{\Delta })\right)}^{-1}{\mathsf{\Delta }}^{+}$. Then, $B^{\#}$ is a GI of $B\iff$ dim $N\left(B\right)=$ dim $N\left(\mathsf{\Delta }\right),$ and codim $R\left(B\right)=$ codim $R\left(\mathsf{\Delta }\right)$.

Define the parameter

$$r=\parallel {\mathsf{\Delta }}^{\#}(\mathsf{\Delta }-E\left(x\right))\parallel ,\forall x\in \mathrm{\Omega }.$$

(9)

We need some estimates.

Lemma 7. 

Let $\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$ and let ${\mathsf{\Delta }}^{\#}\in \mathcal{B}(T_2,T_1)$ be an OI of Δ. Let $E\left(x\right)\in \mathcal{B}(T_1,T_2)$ with $r\in [0,\frac{1}{2})$. Then, ${\mathsf{\Delta }}_1^{\#}$ is a bounded OI of $E\left(x\right)$. Moreover, the following estimates hold.

$$\parallel I-L\parallel \le \frac{r(1-r^m)}{1-r}=b<1,$$

(10)

the operator $L^{-1}\in \mathcal{B}(T_2,T_1)$ and

$$\parallel L^{-1}\parallel \le \frac{1}{1-b}=\overline{b},$$

(11)

where ${\mathsf{\Delta }}_1^{\#},L$ are as defined in the introduction.

Proof. 

The operator ${\mathsf{\Delta }}_1^{\#}$ is a bounded OI of $E\left(x\right)$ by Lemma 2 for $B=E\left(x\right)$. Moreover, we have in turn by the definition of L:

$$\begin{array}{ccc}\hfill \parallel I-L\parallel & =& \parallel H+H^2+·+H^m\parallel \hfill \\ & \le & \parallel H\parallel +{\parallel H\parallel }^2+\cdots +{\parallel H\parallel }^m=\frac{\parallel H\parallel \left(1-{\parallel H\parallel }^m\right)}{1-\parallel H\parallel }\hfill \\ & \le & \frac{r\left(1-r^m\right)}{1-r}=b<1,\hfill \end{array}$$

by the choice of r. It is followed by the Lemma 2 that $H^{-1}\in \mathcal{B}(T_2,T_1)$ and $\parallel H^{-1}\parallel \le \frac{1}{1-b}=\overline{b}$. □

3. Semi-Local Convergence

The convergence of the method (8) is shown using scalar majorizing sequences.

Definition 1. 

Let $\left\{x_n\right\}$ be a sequence in $T_1$. Then, real sequence $\left\{p_n\right\}$ satisfying

$$\parallel x_{n+1}-x_n\parallel \le p_{n+1}-p_n,\forall n\ge 0$$

is called a majorizing sequence for $\left\{x_n\right\}$. If the sequence $\left\{p_n\right\}$ converges, then also $\left\{x_n\right\}$ converges, and for ${\displaystyle x^{\ast }=\underset{n\to \infty }{lim}{x}_n}$ and ${\displaystyle p^{\ast }=\underset{n\to \infty }{lim}{p}_n}$, we have

$$\parallel x^{\ast }-x_n\parallel \le p^{\ast }-p_n.$$

Therefore, the convergence of the sequence $\left\{x_n\right\}$ relates to that of $\left\{p_n\right\}$.

Let $M=[0,+\infty )$.

Some conditions are required in the convergence of (8).

Suppose that

$\left(C_1\right)$

There exists parameters $\kappa \ge 0,r\in \left[0,\frac{1}{2}\right),$ a point $x_0\in \mathrm{\Omega },\mathsf{\Delta }\in \mathcal{B}(T_1,T_2)$ having outer inverse ${\mathsf{\Delta }}^{\#}$ such that $\parallel L{\mathsf{\Delta }}^{\#}\mathrm{{\rm Y}}\left(x_0\right)\parallel \le \kappa$.

$\left(C_2\right)$

There exists function ${\varphi }_0:M\to M$ which is continuous and nondecreasing such that the equation ${\varphi }_0\left(t\right)-1=0$ admits a smallest positive solution $r_0$. Take $M_0=[0,r_0)$.

$\left(C_3\right)$

There exists functions $\varphi :M_0\to M$, ${\varphi }_1:M_0\to M$. Define the real sequence $\left\{s_n\right\}$ for $s_0=0,s_1=\kappa$ as

$$s_{n+1}=s_n+e_n(s_n-s_{n-1}),$$

(12)

where ${\displaystyle e_n=(1+r+\dots +r^m)\left[{\int }_0^1\varphi \left((1-\theta )(s_n-s_{n-1})\right)d\theta +{\varphi }_1\left(s_{n-1}\right)+\overline{b}{r}^{m+1}\right],n=0,1,2,\dots .}$

The sequence $\left\{s_n\right\}$ is proven to be majorizing for $\left\{x_n\right\}$ (see Theorem 1). However, some conditions for this sequence are needed first.

$\left(C_4\right)$

There exists $r\in [\kappa ,r_0)$ such that for all $n=0,1,2,\cdots ,s_n\le r$.

By this condition and (12) that $0\le s_n\le s_{n+1}\le r$ and there exists $s^{\ast }\in [\kappa ,r]$ such that ${\displaystyle \underset{n\to +\infty }{lim}{s}_n=s^{\ast }}$.

The functions ${\phi }_0,\phi$ and ${\phi }_1$ connect to the operators on the method (8).

$\left(C_5\right)$

$\parallel {\mathsf{\Delta }}^{\#}\left({\mathrm{{\rm Y}}}^{\prime }\left(u\right)-\mathsf{\Delta }\right)\parallel \le {\varphi }_0(\parallel u-x_0\parallel )$ for all $u\in \mathrm{\Omega }$. Set ${\mathrm{\Omega }}_0=\mathrm{\Omega }\cap E(x_0,r_0)$, where $E(x_0,r_0)=\{x\in X:\parallel x-x_0\parallel <r_0\}$. We shall also denote by $E[x_0,r_0]$ the closure of $E(x_0,r_0)$.

$\left(C_6\right)$

$r=\parallel {\mathsf{\Delta }}^{\#}\left(\mathsf{\Delta }-E\left(u\right)\right)\parallel <\frac{1}{2},\parallel {\mathsf{\Delta }}^{\#}\left({\mathrm{{\rm Y}}}^{\prime }\left(\tilde{u}\right)-{\mathrm{{\rm Y}}}^{\prime }\left(u\right)\right)\parallel \le \varphi (\parallel \tilde{u}-u\parallel )$ and $\parallel {\mathsf{\Delta }}^{\#}\left({\mathrm{{\rm Y}}}^{\prime }\left(u\right)-E\left(u\right)\right)\parallel \le {\varphi }_1(\parallel u-x_0\parallel )$ for all $u,\tilde{u}\in {\mathrm{\Omega }}_0.$

$\left(C_7\right)$

The equation $e\left(t\right)-1=0$ has a smallest solution in $(0,s^{\ast }]$, where ${\displaystyle e\left(t\right)=(1+r+\cdots +r^m)\left[{\int }_0^1\varphi \left((1-\theta )t\right)d\theta +{\varphi }_1\left(t\right)+\overline{b}{r}^{m+1}\right]}$. Denote such solution by $\overline{s}$ and

$\left(C_8\right)$

$E[x_0,s^{\ast }]\subset \mathrm{\Omega }$.

Next, the convergence is established for (8).

Theorem 1. 

Suppose that the conditions $\left(C_1\right)$–$\left(C_8\right)$ hold. Then, the sequence $\left\{x_n\right\}$ produced by the method (8) converges to a unique solution $x^{\ast }\in E[x_0,s^{\ast }]\cap \left\{x_0+R\left({\mathsf{\Delta }}^{\#}\right)\right\}$ of the equation ${\mathsf{\Delta }}^{\#}\mathrm{{\rm Y}}\left(x_0\right)=0$. Moreover, the following assertion holds

$$\parallel x^{\ast }-x_n\parallel \le s^{\ast }-s_n.$$

(13)

Proof. 

Mathematical induction on n shall establish the estimates

$$\parallel x_{n+1}-x_n\parallel \le s_{n+1}-s_n.$$

(14)

Since

$$\parallel x_1-x_0\parallel =\parallel E{\left(x_0\right)}^{\#}\mathrm{{\rm Y}}\left(x_0\right)\parallel \le \kappa =s_1=s_1-s_0<s^{\ast },$$

the assertion (13) holds if $n=0$ and $x_1\in E(x_0,s^{\ast })$. It follows by (9) (for $x=x_1$) and Lemma 7 that ${\mathsf{\Delta }}_1{\left(x_1\right)}^{\#}$ is an outer inverse of $E\left(x_1\right)$ and $\parallel {\mathsf{\Delta }}_1{\left(x_1\right)}^{\#}A\parallel \le \overline{b}$ and $N\left({\mathsf{\Delta }}_1{\left(x_1\right)}^{\#}\right)=N\left({\mathsf{\Delta }}^{\#}\right)$. Suppose that for $i=1,2,\cdots ,n;\parallel x_i-x_{i-1}\parallel \le s_i-s_{i-1}$ and $N\left({\mathsf{\Delta }}_1{\left(x_{n-1}\right)}^{\#}\right)=N\left({\mathsf{\Delta }}^{\#}\right)$. Then, we have

$$\begin{array}{ccc}\hfill \parallel x_n-x_0\parallel & \le & \parallel x_n-x_{n-1}\parallel +\cdots +\parallel x_1-x_0\parallel \hfill \\ & \le & s_n-s_{n-1}+\cdots +s_1-s_0=s_n<s^{\ast },\hfill \end{array}$$

and $N\left({\mathsf{\Delta }}_1{\left(x_n\right)}^{\#}\right)=N\left({\mathsf{\Delta }}_1{\left(x_{n-1}\right)}^{\#}\right)=N\left({\mathsf{\Delta }}^{\#}\right)$. Hence, by the Lemma 3, it follows that

$${\mathsf{\Delta }}_1{\left(x_n\right)}^{\#}\left(I-E\left(x_{n-1}\right){\mathsf{\Delta }}_1{\left(x_{n-1}\right)}^{\#}\right)=0.$$

Then, by the method (8)

$$\begin{array}{ccc}\hfill \mathrm{{\rm Y}}\left(x_n\right)& =& \mathrm{{\rm Y}}\left(x_n\right)-\mathrm{{\rm Y}}\left(x_{n-1}\right)-\mathsf{\Delta }{L}^{-1}(x_n-x_{n-1})\hfill \\ \hfill & =& \mathrm{{\rm Y}}\left(x_n\right)-\mathrm{{\rm Y}}\left(x_{n-1}\right)-{\mathrm{{\rm Y}}}^{\prime }\left(x_{n-1}\right)(x_n-x_{n-1})\hfill \\ \hfill & & +\left({\mathrm{{\rm Y}}}^{\prime }\left(x_{n-1}\right)-A_{n-1}\right)(x_n-x_{n-1})+\left(A_{n-1}-\mathsf{\Delta }{L}^{-1}\right)(x_n-x_{n-1})\hfill \\ \hfill & =& \mathrm{{\rm Y}}\left(x_n\right)-\mathrm{{\rm Y}}\left(x_{n-1}\right)-{\mathrm{{\rm Y}}}^{\prime }\left(x_{n-1}\right)(x_n-x_{n-1})\hfill \\ \hfill & & +\left({\mathrm{{\rm Y}}}^{\prime }\left(x_n\right)-A_{n-1}-A_{n-1}\right)(x_n-x_{n-1})\hfill \\ \hfill & & +\left(A_{n-1}L-\mathsf{\Delta }\right)L^{-1}(x_n-x_{n-1}).\hfill \end{array}$$

(15)

But, we have by the definition of L

$$\begin{array}{ccc}\hfill A_{n-1}L-\mathsf{\Delta }& =& A_{n-1}(I+H+\cdots +H^m)-\mathsf{\Delta }\hfill \\ & =& A_{n-1}-\mathsf{\Delta }+(-\mathsf{\Delta }+\mathsf{\Delta }+A_{n-1})(I+H+\cdots +H^m)\hfill \\ & =& A_{n-1}-\mathsf{\Delta }+\mathsf{\Delta }(H+\cdots +H^m)-(\mathsf{\Delta }-H)(H+\cdots +H^m)\hfill \\ & =& A_{n-1}-\mathsf{\Delta }+\mathsf{\Delta }H+\mathsf{\Delta }(H^2+\cdots +H^m)\hfill \\ & & -(\mathsf{\Delta }-A_{n-1})(H+\cdots +H^m)\hfill \\ & =& \mathsf{\Delta }(H^2+\cdots +H^m)-\mathsf{\Delta }(\mathsf{\Delta }-A_{n-1})(H+\cdots +H^m).\hfill \end{array}$$

So,

$${\mathsf{\Delta }}^{\#}(A_{n-1}L-\mathsf{\Delta })=-H^{m+1},$$

(16)

where we also used $A_{n-1}-\mathsf{\Delta }+\mathsf{\Delta }H=0$ by the definition of H. Using the induction hypotheses and the conditions $\left(C_5\right),\left(C_6\right)$, (14), (15), (11), we have in turn that

$$\begin{array}{ccc}\hfill \parallel {\mathsf{\Delta }}^{\#}\mathrm{{\rm Y}}\left(x_n\right)\parallel & \le & {\int }_0^1\varphi \left((1-\theta )\parallel x_n-x_{n-1}\parallel \right)d\theta \parallel x_n-x_{n-1}\parallel \hfill \\ \hfill & & +{\varphi }_1(\parallel x_{n-1}-x_0\parallel )\parallel x_n-x_{n-1}\parallel +\overline{b}{r}^{m+1}\parallel x_n-x_{n-1}\parallel \hfill \\ \hfill & \le & \left[{\int }_0^1\left((1-\theta )(s_n-s_{n-1})\right)d\theta +{\varphi }_1\left(s_{n-1}\right)+\overline{b}{r}^{m+1}\right]\hfill \\ \hfill & & \times (s_n-s_{n-1}).\hfill \end{array}$$

(17)

Hence, by (8), Lemma 7 and (17)

$$\begin{array}{ccc}\hfill \parallel x_{n+1}-x_n\parallel & \le & (1+r+\cdots +r^m)[{\int }_0^1\varphi \left((1-\theta )(s_n-s_{n-1})\right)d\theta \hfill \\ & & +{\varphi }_1\left(s_{n-1}\right)+\overline{b}{r}^{m+1}](s_n-s_{n-1})=s_{n+1}-s_n\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel x_{n+1}-x_0\parallel & \le & \parallel x_{n+1}-x_n\parallel +\parallel x_n-x_0\parallel \hfill \\ & \le & s_{n+1}-s_n+s_n-s_0=s_{n+1}<s^{\ast }.\hfill \end{array}$$

The induction is completed. Thus, we have for any n

$$\parallel x_{n+1}-x_n\parallel \le s_{n+1}-s_n,$$

$$\parallel x_n-x_0\parallel \le s_n\le s^{\ast }$$

$$\parallel {\mathsf{\Delta }}^{\#}\left(E\left(x_{n+1}\right)-\mathsf{\Delta }\right)\parallel <1,$$

and

$${\mathsf{\Delta }}_1{\left(x_{n+1}\right)}^{\#}:={\left(I+{\mathsf{\Delta }}^{\#}\left(E\left(x_{n+1}\right)-\mathsf{\Delta }\right)\right)}^{-1}{\mathsf{\Delta }}^{\#}$$

is an OI of $E\left(x_{n+1}\right)$. The sequence $\left\{s_n\right\}$ is complete as convergent and majorizes $\left\{x_n\right\}$. So, the sequence $\left\{x_n\right\}$ is also complete in $T_1.$ Then, it is convergent to a $x^{\ast }\in E[x_0,s^{\ast }]$. By the definition

$${\mathsf{\Delta }}_1{\left(x_n\right)}^{\#}={\left(I+{\mathsf{\Delta }}^{\#}\left(E\left(x_n\right)-A\right)\right)}^{-1}{\mathsf{\Delta }}^{\ast },\mathrm{for}\mathrm{all}n$$

and

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left(I+{\mathsf{\Delta }}^{\#}\left(E\left(x_n\right)-A\right)\right)(x_n-x_{n-1})\hfill \\ \hfill & =& \underset{n\to \infty }{lim}{\mathsf{\Delta }}^{\#}\mathrm{{\rm Y}}\left(x_n\right)={\mathsf{\Delta }}^{\#}\mathrm{{\rm Y}}\left(x^{\ast }\right).\hfill \end{array}$$

(18)

Thus, $x^{\ast }$ solves ${\mathsf{\Delta }}^{\#}\mathrm{{\rm Y}}\left(x\right)=0$. Using Lemma 2, we have $R\left({\mathsf{\Delta }}_1{\left(x_n\right)}^{\#}\right)=R\left({\mathsf{\Delta }}^{\#}\right)$ for all $n=0,1,2,\cdots$. So, $x_{n+1}-x_n=-{\mathsf{\Delta }}_1{\left(x_n\right)}^{\#}\mathrm{{\rm Y}}\left(x_n\right)\in R\left({\mathsf{\Delta }}^{\#}\right)$ and from Lemma 1, we obtain $R\left({\mathsf{\Delta }}^{\#}\right)=R\left({\mathsf{\Delta }}^{\#}\mathsf{\Delta }\right)$, so $x_{n+1}\in x_n+R\left({\mathsf{\Delta }}^{\#}\right)$. Thus, $x_n\in x_0+R\left({\mathsf{\Delta }}^{\#}\right)$ for all n. Suppose that $w\in E[x_0,s^{\ast }]\cap \{x_0+R\left({\mathsf{\Delta }}^{\#}\right)\}$ solves the equation ${\mathsf{\Delta }}^{\#}\mathrm{{\rm Y}}\left(x\right)=0$. Then, we have $w-x^{\ast }\in R\left({\mathsf{\Delta }}^{\#}\right)$ and ${\mathsf{\Delta }}^{\#}E(w-x_n)={\mathsf{\Delta }}^{\#}E(w-x_0)+{\mathsf{\Delta }}^{\#}E(x_n-x_0)=w-x_n$, for all $n=0,1,2,\dots$. Then, (11), as in (16) and using $\left(C_7\right)$

$$\begin{array}{ccc}\hfill \parallel w-x_{n+1}\parallel & =& \parallel w-x_n+{\mathsf{\Delta }}_1{\left(x_n\right)}^{\#}\mathrm{{\rm Y}}\left(x_n\right)-{\mathsf{\Delta }}_1{\left(x_n\right)}^{\#}\mathrm{{\rm Y}}\left(w\right)\parallel \hfill \\ \hfill & \le & (1+r+\cdots +r^m)[{\int }_0^1\varphi \left((1-\theta )\parallel w-x_n\parallel \right)d\theta \hfill \\ \hfill & & +{\varphi }_1(\parallel x_n-x_0\parallel )+\overline{b}{r}^{m+1}]\parallel w-x_n\parallel \hfill \end{array}$$

$$\begin{array}{ccc}\hfill & \le & d\parallel w-x_n\parallel \le d^{n+1}\parallel x_0-w\parallel <\overline{b}\le s^{\ast },\hfill \end{array}$$

(19)

where ${\displaystyle d=(1+r+\cdots +r^m)\left[{\int }_0^1\varphi \left((1-\theta )\parallel w-x_0\parallel \right)d\theta +{\varphi }_1\left(s^{\ast }\right)+\overline{b}{r}^{m+1}\right]\in [0,1)}$. Therefore, we conclude ${\displaystyle x^{\ast }=\underset{n\to +\infty }{lim}{x}_n=w}$. Finally, from (14) and the triangle inequality, we obtain for $j=0,1,2,\dots$

$$\begin{array}{ccc}\hfill \parallel x_{n+j}-x_n\parallel & \le & \parallel x_{n+j}-x_{n+j-1}\parallel +\parallel x_{n+j-1}-x_{n+j-2}\parallel +\cdots +\parallel x_{n+1}-x_n\parallel \hfill \\ & \le & s_{n+j}-s_{n+j-1}+s_{n+j-1}-s_{n+j-2}+\cdots +s_{n+1}-s_n\hfill \\ & =& s_{n+j}-s_n.\hfill \end{array}$$

By letting $j\to +\infty$ in (19), we show the assertion (13). □

Remark 1.

(i) 

The results of the Theorem 1 specialize for the Newton method with OI defined by $x_{n+1}=x_n-{\mathrm{{\rm Y}}}^{\prime }{\left(x_n\right)}^{\#}\mathrm{{\rm Y}}\left(x_n\right)$, for solving Equation (5). Simply, take $E\left(x\right)={\mathrm{{\rm Y}}}^{\prime }\left(x\right)$ and ${\varphi }_1=0$.

(ii) 

Under the conditions $\left(C_1\right)-\left(C_8\right)$, further suppose that the operator ${\left(I+{\mathsf{\Delta }}^{+}\left(\mathsf{\Delta }-E\left(x\right)\right)\right)}^{-1}$$E\left(x\right)$ sends $N\left(A\right)$ to $R\left(A\right)$ provided that for $x\in \mathrm{\Omega }$, the inverse of

$$I+{\mathsf{\Delta }}^{+}\left(E\left(x\right)-\mathsf{\Delta }\right)exists.$$

(20)

Then, by Lemma 4, $E{\left(x_n\right)}^{\#}:={\left(I+{\mathsf{\Delta }}^{+}\left(E\left(x_n\right)-\mathsf{\Delta }\right)\right)}^{-1}{\mathsf{\Delta }}^{+}$ is a GI. Thus, the proof of Theorem 1 establishes the convergence of method (8) for GI.

(iii) 

By the Lemma 6, the condition (20) can be exchanged by rank $\left(E\left(x\right)\right)\le rank\left(E\left(x_0\right)\right)$ for $\mathsf{\Delta }=E\left(x_0\right)$ and if $T_1$ and $T_2$ are finite dimensional. In general Banach spaces, the condition (20) can be switched by the stronger $R\left(E\left(x\right)\right)\subset R\left(E\left(x_0\right)\right)$ (for $\mathsf{\Delta }=E\left(x_0\right)$) (see Lemma 5) or by the conditions of the Lemma 6.

4. Examples

The example considers method (4) with ${\mathrm{{\rm Y}}}^{\prime }{\left(x\right)}^{+}={\mathrm{{\rm Y}}}^{\prime }{\left(x\right)}^{-1}$ for the case $\mathsf{\Delta }=I$, which is independent of $x_0$. It is also compared with method (8) for $\mathsf{\Delta }={\mathrm{{\rm Y}}}^{\prime }\left(x_0\right)$. In this case, the methods (4) and (8) become Newton’s method

$$x_{n+1}=x_n-{\mathrm{{\rm Y}}}^{\prime }{\left(x_n\right)}^{-1}\mathrm{{\rm Y}}\left(x\right){\left(4\right)}^{\prime }$$

and

$$x_{n+1}=x_n-L{\mathrm{{\rm Y}}}^{\prime }{\left(x_0\right)}^{-1}\mathrm{{\rm Y}}\left(x\right){\left(8\right)}^{\prime },$$

respectively.

We shall solve the system

$$\begin{array}{ccc}\hfill F_1(u,v)& =& u-0.1sinu-0.3cosv+0.4\hfill \\ \hfill F_2(u,v)& =& v-0.2cosu+0.1sinv+0.3.\hfill \end{array}$$

If $F=(F_1,F_2).$ Then, we can write

$$\begin{array}{c}\hfill \mathrm{{\rm Y}}\left(z\right)=0forz={(u,v)}^T.\end{array}$$

Consequently, we obtain

$${\mathrm{{\rm Y}}}^{\prime }\left((u,v)\right)=\left[\begin{array}{cc}1-0.1cos\left(u\right)& 0.3sin\left(v\right)\\ 0.2sin\left(u\right)& 0.1cos\left(v\right)+1\end{array}\right].$$

Example 1. 

Method (8′): Set $m=1$ and $\mathsf{\Delta }=I.$ We have that

$$\begin{array}{c}{L}_1\left(x\right)=I+(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right)),\\ P_1\left(x\right)=x-(I+(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right)))\mathrm{{\rm Y}}\left(x\right),\\ x_{j+1}=P_1\left(x_j\right).\end{array}$$

(21)

Example 2. 

Method (8′): Set, $m=2$ and $\mathsf{\Delta }=I.$ It follows that

$$\begin{array}{c}{L}_2\left(x\right)=I+(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^2,\\ P_2\left(x\right)=x-L_2\left(x\right)\mathrm{{\rm Y}}\left(x\right),\\ x_{j+1}=P_2\left(x_j\right).\end{array}$$

(22)

Example 3. 

Method (8′): Set $m=3$ and $\mathsf{\Delta }=I.$ Then, we have

$$\begin{array}{c}{L}_3\left(x\right)=I+(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^2+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^3,\\ P_3\left(x\right)=x-L_3\left(x\right)\mathrm{{\rm Y}}\left(x\right),\\ x_{j+1}=P_3\left(x_j\right).\end{array}$$

(23)

Example 4. 

Method (8′): Set $m=4$ and $\mathsf{\Delta }=I.$ Then, we have

$$\begin{array}{c}{L}_4\left(x\right)=I+(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^2+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^3+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^4,\\ P_4\left(x\right)=x-L_4\left(x\right)\mathrm{{\rm Y}}\left(x\right),\\ x_{j+1}=P_4\left(x_j\right).\end{array}$$

(24)

Example 5. 

Method (8′): Set $m=5$ and $\mathsf{\Delta }=I.$ Then, we have

$$\begin{array}{c}{L}_5\left(x\right)=I+(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^2+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^3+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^4+{(I-{\mathrm{{\rm Y}}}^{\prime }\left(x\right))}^5,\\ P_5\left(x\right)=x-L_5\left(x\right)\mathrm{{\rm Y}}\left(x\right),\\ x_{j+1}=P_5\left(x_j\right).\end{array}$$

(25)

Example 6. 

Method (8′): Set $m=\overline{1,5}$ and $\mathsf{\Delta }={\mathrm{{\rm Y}}}^{\prime }\left(x_0\right).$ It follows that

$$\begin{array}{c}{x}_{n+1}=x_n-L{\mathsf{\Delta }}^{-1}\mathrm{{\rm Y}}\left(x_n\right),\\ H={\mathsf{\Delta }}^{-1}(\mathsf{\Delta }-{\mathrm{{\rm Y}}}^{\prime }\left(x\right)),\\ L=I+\sum _{j=1}^m{H}^j.\end{array}$$

(26)

Definition 2. 

Let $\left\{x_n\right\}$ be a sequence. Then, the computational order of convergence (COC) is for ${\theta }_n=x_n-x^{\ast }$ [4]

$${\overline{h}}_n=\frac{ln|{\theta }_{n+1}/{\theta }_n|}{ln|{\theta }_n/{\theta }_{n-1}|}.$$

Definition 3. 

Let $\left\{x_n\right\}$ be a sequence. Then, the approximate computational order of convergence (ACOC) is for ${\widehat{\theta }}_n=x_n-x_{n-1}$

$${\widehat{h}}_n=\frac{ln|{\widehat{\theta }}_{n+1}/{\widehat{\theta }}_n|}{ln|{\widehat{\theta }}_n/{\widehat{\theta }}_{n-1}|}.$$

The Table 1, Table 2, Table 3, Table 4 and Table 5 demonstrate that the cheaper-to-implement method (8) is behaving the same as Newton’s method for a large enough $m.$

Table 1. Iterations to obtain error tolerance $\epsilon ={10}^{-9}$ for initial point $x_0=(1,1),$ where $\parallel I-{\mathrm{{\rm Y}}}^{\prime }\left(x_0\right)\parallel =0.3129<1$.

Method	Iterations	CPU Time	Method	Iterations	CPU Time
(4)’ Newton	4	$10.82\times {10}^{-6}$	(4)’ Newton	4	$10.82\times {10}^{-6}$
(21), $m=1$	6	$6.94\times {10}^{-6}$	(26), $m=1$	8	$13.063\times {10}^{-6}$
(22), $m=2$	5	$6.922\times {10}^{-6}$	(26), $m=2$	6	$12.255\times {10}^{-6}$
(23), $m=3$	4	$5.946\times {10}^{-6}$	(26), $m=3$	5	$9.089\times {10}^{-6}$
(24), $m=4$	4	$6.324\times {10}^{-6}$	(26), $m=4$	5	$9.198\times {10}^{-6}$
(25), $m=5$	4	$6.836\times {10}^{-6}$	(26), $m=5$	4	$8.743\times {10}^{-6}$

Table 2. Iterations to obtain error tolerance of $\epsilon ={10}^{-9}$ for initial point $x_0=(0,0),$ where $\parallel I-{\mathrm{{\rm Y}}}^{\prime }\left(x_0\right)\parallel =0.1414<1$.

Method	Iterations	CPU Time	Method	Iterations	CPU Time
(4)’ Newton	3	$7.215\times {10}^{-6}$	(4)’ Newton	3	$7.215\times {10}^{-6}$
(21), $m=1$	5	$5.122\times {10}^{-6}$	(26), $m=1$	3	$4.256\times {10}^{-6}$
(22), $m=2$	4	$5.8\times {10}^{-6}$	(26), $m=2$	3	$5.862\times {10}^{-6}$
(23), $m=3$	3	$5.503\times {10}^{-6}$	(26), $m=3$	3	$5.575\times {10}^{-6}$
(24), $m=4$	3	$5.771\times {10}^{-6}$	(26), $m=4$	3	$5.825\times {10}^{-6}$
(25), $m=5$	3	$5.895\times {10}^{-6}$	(26), $m=5$	3	$5.973\times {10}^{-6}$

Table 3. Iterations to obtain error tolerance of $\epsilon ={10}^{-9}$, for $x_0=(-15,-15),$ where $\parallel I-{\mathrm{{\rm Y}}}^{\prime }\left(x_0\right)\parallel =0.257<1$.

Method	Iterations	CPU Time	Method	Iterations	CPU Time
(4)’ Newton	5	$13.454\times {10}^{-6}$	(4)’ Newton	5	$13.454\times {10}^{-6}$
(21), $m=1$	7	$7.184\times {10}^{-6}$	(26), $m=1$	9	$12.015\times {10}^{-6}$
(22), $m=2$	5	$10.195\times {10}^{-6}$	(26), $m=2$	7	$13.245\times {10}^{-6}$
(23), $m=3$	5	$7.352\times {10}^{-6}$	(26), $m=3$	6	$9.378\times {10}^{-6}$
(24), $m=4$	5	$7.883\times {10}^{-6}$	(26), $m=4$	6	$9.829\times {10}^{-6}$
(25), $m=5$	5	$8.636\times {10}^{-6}$	(26), $m=5$	5	$8.936\times {10}^{-6}$

Table 4. Iterations to obtain error tolerance of $\epsilon ={10}^{-12}$, for $x_0=(-15,-15),$ where $\parallel I-{\mathrm{{\rm Y}}}^{\prime }\left(x_0\right)\parallel =0.257<1$.

Method	Iterations	CPU Time	Method	Iterations	CPU Time
(4)’ Newton	7	$16.719\times {10}^{-6}$	(4)’ Newton	7	$16.719\times {10}^{-6}$
(21), $m=1$	8	$15.677\times {10}^{-6}$	(26), $m=1$	12	$24.742\times {10}^{-6}$
(22), $m=2$	7	$20.052\times {10}^{-6}$	(26), $m=2$	8	$27.178\times {10}^{-6}$
(23), $m=3$	7	$14.711\times {10}^{-6}$	(26), $m=3$	8	$20.43\times {10}^{-6}$
(24), $m=4$	7	$15.357\times {10}^{-6}$	(26), $m=4$	7	$19.2\times {10}^{-6}$
(25), $m=5$	7	$16.49\times {10}^{-6}$	(26), $m=5$	7	$17.538\times {10}^{-6}$

Table 5. COC versus ACOC with $x_0=(-15,-15)$, $\epsilon ={10}^{-12}$.

Method	COC	ACOC
(2)   Newton	1.8624	1.9697
(21),   $m=1$	0.863	1
(22),   $m=2$	0.2695	1.0438
(23),   $m=3$	1.9714	2.3569
(24),   $m=4$	1.8354	1.9453
(25),   $m=5$	1.8642	1.9661
(26),   $m=1$	0.9065	1.0118
(26),   $m=2$	0.5912	0.999
(26),   $m=3$	0.7321	0.9926
(26),   $m=4$	1.933	2.0151
(26),   $m=5$	1.8679	1.9578

5. Conclusions

We developed a semi-local Kantorovich-like analysis of Newton-type methods for solving singular nonlinear operator equations using outer or generalized inverses. These methods do not use inverses as in earlier studies but a sum of operators. This sum converges to the inverse and makes the implementation of these methods easier than the ones using inverses. The analysis of the methods relies on the concept of generalized continuity for the operators involved and majorizing sequences. Examples complement the theory. Due to its generality, this article’s technique can be applied on other method with inverses along the same lines [6,14,19,32,39,43,45,46,47,48,49]. It is worth noting that the method (8) should be used for sufficiently small $m.$ Otherwise, if m is very large, it may be as expensive to implement as method (4).